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Probability -- traffic lights (1. R=0.4, 2. R=0.5, 2x G=0.2)

  1. Mar 7, 2016 #1

    TheBlackAdder

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    There are 2 traffic lights. Bart knows the probability of stopping at the first light is 40%, stopping at the second light is 50% and the probability of stopping at no lights is 20%.
    1. What is the probability of stopping at both lights?
    2. What is the probability of stopping for at least one light?
    My attempt:
    yu868Zl.jpg

    Apparently the chance is 10% and not 20% of stopping at boh lights. I thought they were independent but the opposite is true. So I used the formula for dependent probability. Now I'm trying to figure out how I can find P(B=R | A=R)
     
    Last edited: Mar 7, 2016
  2. jcsd
  3. Mar 7, 2016 #2

    BvU

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    But already earlier on you have a problem: case G G would be 0.6 * 0.5 and given is that it's only 0.2.

    That tells you something about case G R !

    The other given , 2nd light 0.5 / 0.5 on average, can then only be explained if case R G is a lot more likely than 0.4 * 0.5
     
  4. Mar 7, 2016 #3

    mfb

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    They are dependent, yes.
    You don't need formulas for this. Just consider all four options:
    (A) stop at both
    (B) stop at first but not at second
    (C) stop at second but not at first
    (D) both green

    Those four options sum to a probability of 1, and you can set up three equations with the given numbers. That is sufficient to find all probabilities.
     
  5. Mar 7, 2016 #4

    TheBlackAdder

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    These are some serious cardinal mistakes I'm making. Thank you.
     
  6. Mar 7, 2016 #5

    TheBlackAdder

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    I understand my mistakes. But I really don't see how I can use the formula. To calculate the conditional probability P(B=R | A=R) I need the probability of P(B=R ∩ A=R) but that is what I'm looking for in the first place.

    Our lecturer did this: P(X={red}) = 40%, P(Y={red}) = 50% (probability of red at the second light)
    P(X=Red ∩ Y=Red) = P(X=R | Y=R)
    P(X=Red ∩ Y=Red) = P(X=R) + P(Y=R) - P(X=R ∪ Y=R) = 40% + 50% - 80%
    P(X=Red ∩ Y=Red) = 10%

    I don't see how he got to P(X=R ∪ Y=R) = 80%

    m2ntrGn.jpg
     
  7. Mar 7, 2016 #6

    BvU

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    OK, you picked up GG = 0.2 and hence GR = 0.4.
    To end up with *G = 0.5 you need RG = 0.3 - that's one way.
    To end up with *R = 0.5 you need RR = 0.1 - that's an even more direct way.
    *G is the chance you need to stop at light #2.

    Teacher finds chance that one or both are red = 100% - chance that both are green (which is given). That explains the 80%.
    But I can't follow the logic in
    can you ?
     
  8. Mar 7, 2016 #7

    TheBlackAdder

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    Oh I think I somewhat start to see what he did.
    P(X=R) + P(Y=R) = 90% = The probability the first one is red (R) + the probability the second one is red (GR + RR)
    -P(X=R ∪ Y=R) = The first light is red (R) or the second light is red (GR) or both are red (RR) ~ which equals the second question: the probability of at least one light being red = 80%

    After writing this I don't see anymore. I really want to grasp this, but can't seem to find the logic. So my answer is no @BvU :(
     
  9. Mar 7, 2016 #8

    mfb

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    You don't need conditional probabilities at all.
    If they confuse you, why do you try to keep using them?
     
  10. Mar 7, 2016 #9

    TheBlackAdder

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    Because we need to learn to write it in a formula. I'll figure it out tomorrow. Will update this thread.
     
  11. Mar 7, 2016 #10

    mfb

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    Then I suggest to start with the approach I posted and to calculate conditional probabilities afterwards (unnecessary, but...).
     
  12. Mar 7, 2016 #11

    Ray Vickson

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    Let R1 = {first is red (stop)}, R2 = {second is red (stop)}. You are given ##P(R_1) = 0.4, P(R_2) = 0.5## and ##P(\text{not}(R_1 \cup R_2)) = 0.2##. Thus, ##P(R_1 \cup R_2) = 1-0.2 = 0.8##. The inclusion-exclusion principle says that
    [tex]P(R_1 \cup R_2) = P(R_1) + P(R_2) - P(R_1 \cap R_2), [/tex]
    so ##P(R_1 \cap R_2) = 0.4 + 0.5 - 0.8 = 0.1##.
     
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